Question

Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.

Short answer

The solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(4.95 \times 10^{-11} M\).

Step-by-step solution

Write Balanced Chemical Equation

For the dissolution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in water, we can write the balanced chemical equation as follows: \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ca^{2+}}(aq) + 2\mathrm{PO^{3-}_{4}}(aq)\]

Write the Solubility Product Expression

Given the balanced chemical equation, we can now write the solubility product expression \(K_{sp}\) for the reaction. \[K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO^{3-}_{4}}]^2\]

Set Up the ICE Table

To find the equilibrium concentrations, we can set up an ICE (Initial, Change, Equilibrium) table for the reaction: \[\begin{array}{c|ccc} & \mathrm{Ca}^{2+} & \mathrm{PO}_{4}^{3-} \\ \hline \text{Initial} & 0 & 0.20 \\ \text{Change} & +3s & +2s \\ \text{Equilibrium} & 3s & 0.20 + 2s \end{array}\] Where s is solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Substituting the Equilibrium Concentrations into \(K_{sp}\) Value

Substituting the equilibrium concentrations into the solubility product expression, and using the given \(K_{sp} = 1.3 \times 10^{-32}\): \[1.3 \times 10^{-32} = (3s)^3(0.20 + 2s)^2\]

Solve for Solubility

Since \(K_{sp}\) is a very small number, we can approximate that \(0.20 + 2s \approx 0.20\). Therefore, the equation becomes: \[1.3 \times 10^{-32} = (3s)^3(0.20)^2\] Solve for s: \(s \approx 4.95 \times 10^{-11} \mathrm{M}\) The solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(4.95 \times 10^{-11} \mathrm{M}\).

Key concepts

Solubility Product (Ksp)

The solubility product, denoted as \( K_{sp} \), is a constant that provides insight into the solubility of a sparingly soluble salt in a solution. It is essentially the equilibrium constant specific for a solvation reaction, expressing the product of the concentrations of the ions that the substance dissociates into when in a saturated solution. For example, in the dissolution of calcium phosphate \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \):

• The chemical reaction is
  \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ca^{2+}}(aq) + 2\mathrm{PO^{3-}_{4}}(aq)\]
• The expression for \( K_{sp} \) for this reaction is
  \[K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO^{3-}_{4}}]^2\]

In practice, a smaller \( K_{sp} \) value means the compound is less soluble. This plays a significant role in predicting whether a precipitate will form in a solution. If the ionic concentrations product exceeds \( K_{sp} \), the solution forms a precipitate.

ICE Table

An ICE table is a simple yet powerful tool used in chemistry to organize and calculate the changes in concentrations of species participating in a chemical reaction as they achieve equilibrium. ICE stands for Initial, Change, Equilibrium. Let's break down how it works for this example:

• Initial: Begin with the initial concentrations of ions before any dissolution. In our case, \([\mathrm{Ca^{2+}}] = 0\) and \([\mathrm{PO^{3-}_{4}}] = 0.20\) M from the sodium phosphate.
• Change: This represents the change in concentration as the system approaches equilibrium. For every "s" moles of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) that dissolves, \([\mathrm{Ca^{2+}}]\) increases by "3s" and \([\mathrm{PO^{3-}_{4}}]\) by "2s".
• Equilibrium: The total concentration at equilibrium is found by combining initial and change. Hence, \([\mathrm{Ca^{2+}}] = 3s\) and \([\mathrm{PO^{3-}_{4}}] = 0.20 + 2s\).

The ICE table simplifies complex equilibrium problems, allowing for systematic solving to find unknown concentrations.

Chemical Equilibrium

Chemical equilibrium exists when a chemical reaction's forward and backward reactions occur at an equal rate, leading to constant concentrations of reactants and products over time. Although it may seem reactions have stopped, they are dynamically balanced.For dissolution, equilibrium is reached when the rate of the solid dissolving equals the rate of its recrystallization from the solution. In our example of \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \):

• Dynamic Balance: At equilibrium, soluble ions of calcium and phosphate form and dissolve from the solid at the same rate.
• Concentration Stability: The concentrations of \([\mathrm{Ca^{2+}}]\) and \([\mathrm{PO^{3-}_{4}}]\) remain steady, signifying equilibrium.

Calculating \( K_{sp} \) and using an ICE table assists in predicting these equilibrium concentrations efficiently. This knowledge is crucial not only in theoretical calculations but also in practical applications like drug formulation and wastewater treatment, where solubility matters.